Question Number 226910 by Estevao last updated on 18/Dec/25
Answered by Estevao last updated on 18/Dec/25
$$>{Ashed}\:{Area} \\ $$
Answered by fantastic2 last updated on 19/Dec/25
Commented by fantastic2 last updated on 19/Dec/25
$${radius}\:{of}\:{big}\:{semicircle}=\mathrm{6}+\mathrm{3}=\mathrm{9}{cm} \\ $$$${R}^{\mathrm{2}} +\left(\mathrm{18}−\mathrm{6}−{r}\right)^{\mathrm{2}} =\left({R}+{r}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{R}^{\mathrm{2}} +\mathrm{144}−\mathrm{24}{r}+{r}^{\mathrm{2}} ={R}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}{Rr} \\ $$$$\Rightarrow\mathrm{144}=\mathrm{2}{Rr}+\mathrm{24}{r} \\ $$$${blue}\:{length}=\mathrm{9}{cm} \\ $$$${green}\:{line}\:{is}\:{going}\:{through}\:{the} \\ $$$${midpoint}\:{of}\:{the}\:{blue}\:{semicircle} \\ $$$$\therefore{its}\bot{on}\:{R} \\ $$$${green}=\sqrt{\mathrm{81}−{R}^{\mathrm{2}} } \\ $$$${R}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} =\left(\sqrt{\mathrm{81}−{R}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{R}^{\mathrm{2}} =\mathrm{72}\Rightarrow{R}=\mathrm{6}{cm} \\ $$$$\mathrm{144}=\mathrm{2}{Rr}+\mathrm{24}{r} \\ $$$$\Rightarrow\mathrm{144}=\mathrm{36}{r}\Rightarrow{r}=\mathrm{4}{cm} \\ $$$${shaded}\:{blue}\:{area}=\frac{\mathrm{1}}{\mathrm{2}}\left(\pi\mathrm{9}^{\mathrm{2}} −\pi\mathrm{6}^{\mathrm{2}} −\pi\mathrm{4}^{\mathrm{2}} \right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{29}\pi=\mathrm{14}.\mathrm{5}\pi{cm}^{\mathrm{2}} \\ $$